为什么在构造函数调用之前初始化实例变量？

Question

public abstract class UCMService{
    private String service;     

    protected DataMap dataMap = new DataMap(); 

    protected class DataMap extends HashMap<String,String> {

        private static final long serialVersionUID = 4014308857539190977L;

        public DataMap(){
            System.out.println("11111");
            put("IdcService",service);
        }
    }

    public UCMService(String service){
        System.out.println("2222");
        this.service = service;
    }
}

Answer 1

这是因为在编译时，编译器会将您在声明位置所做的每个初始化移动到类的每个构造函数。因此，UCMService类的构造函数被有效地编译为：

public UCMService(String service){
    super();  // First compiler adds a super() to chain to super class constructor
    dataMap = new DataMap();   // Compiler moves the initialization here (right after `super()`)
    System.out.println("2222");
    this.service = service;
}

因此，显然datamap()构造函数是在UCMService类的print语句之前执行的。类似地，如果您的UCMService类中还有任何构造函数，那么初始化将移到所有构造函数中。

让我们看看一个简单类的字节代码：

class Demo {
    private String str = "rohit";

    Demo() {
        System.out.println("Hello");
    }
}

Demo();
    Code:
       0: aload_0
       1: invokespecial #1   // Method java/lang/Object."<init>":()V
       4: aload_0
       5: ldc           #2   // String rohit
       7: putfield      #3   // Field str:Ljava/lang/String;
      10: getstatic     #4   // Field java/lang/System.out:Ljava/io/PrintStream;
      13: ldc           #5   // String Hello
      15: invokevirtual #6   // Method java/io/PrintStream.println:(Ljava/lang/String;)V
      18: return